any four point expected: 1/2 each point: 2 marks)

MAHARASHTRA STATE BOARAD OF TECHNICAL EDUCATIOD (Autonomous)

(ISO/IEC-27001-2005 Certified)

SUMMER– 2018 Examinations

Subject Code: 17507 Model Answer of 42

Important suggestions to examiners:

1) The answers should be examined by key words and not as word-to-word as given in the model answer scheme.

2) The model answer and the answer written by candidate may vary but the examiner may try to assess the understanding level of the candidate.

3) The language errors such as grammatical, spelling errors should not be given more importance. (Not applicable for subject English and communication skills)

4) While assessing figures, examiner may give credit for principle components indicated in a figure. The figures drawn by candidate and model answer may vary. The examiner may give credit for any equivalent figure drawn.

5) Credits may be given step wise for numerical problems. In some cases, the assumed constant values may vary and there may be some difference in the candidate’s answers and model answer.

6) In case some questions credit may be given by judgment on part of examiner of relevant answer based on candidate understands.

7) For programming language papers, credit may be given to any other program based on equivalent concept.

Q.1 A) Attempt any THREE : 3 x 4 = 12 Marks

a) Define electric drive. List at least four advantages of electric drive. Ans: Electric Drive: ( 2 Marks)

It is a machine which gives mechanical power. e.g. drives employing electric motors are known as electric drives.

Following advantages of electric drive: ( Any Four point Expected: 1/2 each point: 2 Marks)

1. It is more economical.

2. It is more clean.

3. No air pollution.

4. It occupies less space.

5. It requires less maintenance.

6. Easy to start and control.

7. It can be remote controlled.

8. It is more flexible.

9. Its operating characteristics can be modified.

10. No standby losses.

11. High efficiency.

12. No fuel storage and transportation cost.





13. Less maintenance cost.

14. It has long life.

15. It is reliable source of drive.

b) State the causes of failure of heating element. Ans: Following of the different causes of failure of heating element:

( Any Four causes expected: 1 Mark each)

i) Formation of hot spot:

Hot spot on heating element is the point which is at higher temperature than remaining heating element portion. So there is possibility of breaking of heating element at hot spot.

ii) Due to oxidization:

At high temperature material gets oxidized which may cause failure of heating element.

iii) Due to corrosion:

If heating element is directly exposed to chemical fumes then there is possibility of rusting of heating element which causes failure of heating element.

iv) Mechanical Failure:

Measure heating element alloy contain iron which is brittle. Due to frequent heating &

cooling of heating element, it may break (fail) due to small mechanical injury also.

c) Suggest suitable electric drive for following application :(i) Paper mills (ii) Stone crusher (iii) Textile mill and (iv) Electric traction

Ans: ( Each Suitable Any one Drive suggestion: 1 Mark each)

S.No Application Suitable electric drive

i) Paper mills Slip-Ring Induction Motor, Synchronous Motor

ii) Stone crusher A.C. Series Motor, Slip-Ring Induction Motor

iii) Textile mill Squirrel Cage Induction Motor

iv) Electric traction D.C. Series Motor, 1 Phase Slip-Ring Induction

Motor





d)

Draw the curve and estimate suitable H.P. of motor having following duty cycle : (i) Rising load from 200 to 400 HP 4 minutes (ii) Uniform load of 300 HP 2 minutes (iii) Regenerative braking from 50 to zero H.P. — 1 minute (iv) Idle for 1 minute

Ans: (When Final answer of Numerical is correct Give Full Marks & if final answer is wrong give stepwise marks)

i) Load rising from 200 to 400 HP :- 4 min

ii) Uniform load of 300 HP :- 2 min iii) Regenerative braking from 50 to zero : 1 min

iv) idle for : 1 min

or Equivalent fig----------------(1/2 Mark)

T

tHtHtHHHHHP 3

242

231

2221

21 3

13

1 ---------------------- (1/2 Mark)

Where, T = t1 + t2 + t3 + t4

T = 4 + 2 + 1 + 1

T = 8 min.-------------------------------------------------------------------- (1 Mark)

8

tH31tHtHHHH3

1HP

32

422

312

2212

1

(1/2Mark)

8

15031230044004002002003

1 2222 HP ------------- (1/2 Mark)

241662500

HP

HPHP 263 -------------------------------------Answer------------------ (1 Marks)

Nearest standard rating of motor is to be selected.





Q.1B) Attempt any ONE of the following : 6 Marks a) Describe any six factors governing selection of a motor for a particular application.

Ans: (Any Six Points From The Following Or Equivalent Points Are Expected 1 Mark To Each Point, Total 6 Marks)

Following Factors governing / or are considered while selecting electric drive (Motor) for particular application:

1. Nature of supply: Whether supply available is AC, Pure DC Or Rectified DC

2. Nature of Drive (Motor):

Whether motor is used to drive (run)

Individual machine

OR group of machines.

3. Nature of load:

Whether load required light or heavy starting torque

OR load having high inertia, requirehigh starting torque for long duration.

OR Whether load torque increases with speed (T N)

OR decreases with speed (T N1 )

OR remains constant with speed (T = N)

OR increases with square of speed (T N2)

4. Electric Characteristics of drive:

Starting,

Running,

Speed control

and braking characteristics

of electric drive should be studied and it should be matched with load requirements(i.e. machine).

5. Size and rating of motor:

Whether motor is short time running

OR continuously running

OR intermittently running

OR used for variable load cycle.





Whether overload capacity, pull out torque is sufficient.

6. Mechanical Considerations:

Types of enclosure,

Types of bearing,

Transmission of mechanical power,

Noise

and load equalization

7. Cost: Capital, Running Maintenance cost should be less.

b) State the factors to be considered for selection of shape and size of the car of elevator. Ans: (Any four points are Expected 1.5 Marks to each Total 6 Marks)

The size and shape of elevator car depends on following factors: i) No. of passenger to be carried: While selecting the size of car it is a usual practice to allow.

A Space of 2 Sq.fit/ person.

Average weight of passenger is assumed 68 kg/person.

Thus the maximum load capacity of elevator is considered 34 kg/sq.ft

There should be wide frontage and shallow depth

ii) Limitation in the building design:

Shape of elevator depends on space available in building.

iii) Type of building

iv) Application of elevator

Q.2 Attempt any FOUR : 4 x 4 = 16 Marks

a) Define load equalisation for electric motors. Explain how it is obtained for electric motors. Ans: Define load equalization for electric Motor:

( Meaning : 2 Mark, Figure: 1 Mark & explanation: 1 Mark) There are many types of load which are fluctuating in nature e.g. wood cutting m/c, Rolling

mill. Etc. For such type of loads, load equalization is necessary to draw the constant power from





supply. Because,

When there is sudden load on motor, it will draw more current from supply at start to meet

additional power demand. Due to this heavy current there is large voltage drop in supply system.

This will affect electrical instrument, equipment, m/c, other consumer etc. which are connected

across same supply line.

Also to withstand heavy current, size of input cable increases so cost of cable increases,

Hence it is necessary to smooth out load fluctuations on motor.

The process of smoothing out load fluctuation is called load equalization.

Diagram of Load Equalization:

b) Define : (i) Continuous loading, (ii) Short time loading, (iii) Long time (intermittent) loading, (iv) Continuous operation with short time loading.

Ans: ( Each Definition 1 Mark , Total 4 Marks, Graphical Figure Not expected) (i) Continuous loading:-

or eqivalent figure

This is an output which a motor can deliver continuously without exceeding the permissible

temperature limit.

It can deliver 25% over load for two hours without rise in temperature.





(ii) Short time loading:-

or eqivalent figure

In short time loading motor is operated for short time continuously without exceeding the

permissible temperature limit.e.g. 15min., 20min., 30min. etc than it is made OFF This OFF

load interval is sufficient to cool the motor temperature to its normal value.

(iii) Long time (intermittent) loading:-

or eqivalent figure

Explanation :-In this case motor is operated continuously for long time and interval between two

load is not OFF- load but motor runs at no load for short time. So temperature of

drive continuously increases.

(iv) Continuous operation with short time loading:-

or eqivalent figure

Explanation :-In this case motor is operated continuously for short time and interval between two load

is not OFF- load but motor runs at no load for long time. So temperature of drive continuously

increases. So temperature rise is more than short-time loading.





c) State the principle of induction heating. Write four applications of induction heating. Ans: Principle of Induction heating: ( 2 Mark)

The basic principle of induction heating is that, supply is given to primary winding of furnace

transformer & heat is produced in the secondary (charge) due to electromagnetic action.

OR

Principle of Induction heating:

It is based on principle of transformer. In this type primary winding is as usual which is

wound around one limb of magnetic core but secondary winding is actually charge which is to be

melted is kept in crucible.

When AC Supply is given to primary winding current flows through primary winding

which creates alternating flux in magnetic core this flux links to the secondary winding i.e. charge

through magnetic core. Hence according to faraday’s law of electromagnetic induction emf will

be induced in secondary winding, that is in the charge. As charge forms a close circuit

(secondary) heavy current flows through charge this current is responsible to produce heat in

charge due to I2R losses. This heat is utilized to melt the charge.

Where, R = Resistance of charge & I secondary current

Following are applications of induction heating:

(Any Four point expected: 1/2 each, Total 2 Marks) 1. Melting of steel and non ferrous metals at temperatures up to 1500 °C.

2. Heating for forging to temperatures up to 1250 °C.

3. Annealing and normalizing of metals after cold forming using temperatures in the range of 750 –

950 °C.

4. Surface hardening of steel and cast iron work pieces at temperatures from 850 – 930 °C

(tempering 200-300 °C)

5. Soft and hard soldering at temperatures up to 1100 °C,

6. Moreover, special applications such as heating for sticking, sintering





d) State the principle and nature of supply used for eddy current heating. State the advantages and disadvantages of eddy current heating.

Ans: Figure of Eddy Current Heating:- (1 Mark)

or Equivalent fig.

Principle:- (1 Mark) Heat produced eddy current loss B2 f2

Depth of penetration of heat 1F

The job which is to be heated is wound by coil as shown in figure.

Supply of high voltage (10KV) & high frequency (10-40 KHz) is given to coil which induces

eddy current in job according to Faraday’s law of Electromagnetic induction & these eddy currents

are responsible to produce heat in job itself due to eddy current loss.

In high frequency eddy current heating the phenomenon of skin effect plays an important role.

Skin effect at high frequency is more pronounced (effective). Due to this surface of job is

more heated as compared to its depth.

Nature of supply used for eddy current heating: (1 Mark)

High voltage (10KV)

High frequency (10-40 KHz)

Advantages eddy current heating:- (Any one point expected) (1/2 Mark) 1) No heat transfer loss as heat is produced in job itself. So it has high efficiency.

2) As heat is produced in job itself so time required for heating is less. For e.g. in some cases

operating time taken for heating is of only one second.





3) By simply controlling frequency, we can control temperature accurately.

4) By simply controlling frequency, depth of penetration of heat can be controlled easily.

5) Very thin material surface can be heated easily.

6) Operation is simple & automatic.

7) For heating low attention is required.

8) Heating can be taken place in vacuum or other special atmospheric condition where other

methods are not possible.

9) It is clean and convenient method.

Disadvantages of eddy current heating:- (1/2 Mark)

1) High initial cost because of high voltage high frequency supply equipment is required.

e) Compare Single phase 25 kV AC and 1500 V DC track electrification.

Ans: (Any Four point expected: 1 Mark each, Total 4 Marks)

S.No Points Single phase 25 kV AC track Electrification

1500 V DC track electrification

1. Supply given to O/H condition

1-ph, 25KV, AC 50 Hz 600/750V-Tromways 1500/3000V urban/suburban

2. Type of drive used 1-ph, AC series motor DC series motor for tramways. DC compound motor

3. Weight of traction motor 1.5 times more than d.c. series motor.

1.5 times less than a.c series motor

4. Starting torque Less starting torque than d.c series motor

High starting torque

5. Accln and retardation Less than d.c series motor High 6. Overload capacity Less than d.c series motor High 7. Method of speed control Simple and smooth Limited, except chopper

method 8. Maintenance cost of traction

motor More Less

9. Starting Efficiency More Less 10. Ridding quality Less, better than d.c. Smooth (Better) 11. Insulation cost High Low 12. Cross section of conductor Less More 13. Design of supporting

structure light Heavy





14. Distance between two substation

More Less

15. No. of substation required for same track distance.

Less More

16. Size (capacity) of traction substation

More Less

17. Capital & maintenance cost of substation

Less More

18. Cost track electrification for same track distance

Less More

19. Applications Main line services Urban and suburban area

f) State the various types of welding. Ans: (Total 4 Marks)

i) Resistance Welding:-

1) Spot welding

2) Seam welding

3) Projection Welding

4) Butt Welding- i) Simple butt welding

ii) Flash butt welding

ii) Arc welding:-

1) Carbon Arc Welding: a) shielded welding b) unshielded welding

2) Metal Arc Welding: a) shielded welding b) unshielded welding

Q.3 Attempt any TWO : 2 x 8 = 16 Marks a) i) (i) State advantages and disadvantages of electric braking over mechanical braking.

Ans:

(Any Four Points From The Following Or Equivalent Points Are Expected For Advantages 1/2 Mark To Each Point, Total 2 Marks & For Disadvantages 1/2 Mark To Each Point, Total 2 Marks, Total 4 Marks) Following are the advantages & disadvantages of electrical braking over mechanical braking system.

Advantages: (Any Four Points From The Following Or Equivalent Points Are Expected) 1. It is most reliable braking system.

2. Breaking actuation time is small as higher value of braking retardation is obtained.

3. Electrical braking is smooth & gradual.





4. Life of electrical braking system is more.

5. There is less wear & tear of brake shoes, break block etc. so there is less maintenance cost.

6. Higher speeds are possible even when train is going down the gradient, as breaking system is

reliable.

7. Trains having heavy loads can be stopped even when train going up the gradient.

8. Higher speeds of train is possible as braking system is reliable so pay load capacity increases.

9. In case of electric regenerative braking we can utilize 60 to 80% of kinetic energy to generate

electricity which is not possible with mechanical braking.

Disadvantages: (Any Four Points From The Following Or Equivalent Points Are Expected)

1. In addition to electrical braking there must be arrangement of mechanical braking for final

stop.

2. Special arrangement of circuit and complication makes electrical braking system costly.

3. Operation in substation becomes complicated at the time of regenerative breaking when

generated energy is surplus.

4. Initial cost is more due to other control equipments & circuitry.

a) ii) (ii) State any eight advantages of electric heating. Ans: Advantages of Electric heating:

( Any Four Advantages expected : 1 Mark each, Total 4 Marks) 1. It can be put into service immediately.

2. No standby losses.

3. High efficiency.

4. More economical than other conventional types of heating system.

5. Easy to operate and control.

6. No air pollution.

7. System is clean, as there is no waste produced.

8. No fuel transportation cost.

9. No space is required for storage of fuel and waste.





10. Noiseless operation.

11. Uniform heating is possible; heating at particular point is also possible.

12. Dielectric material can be heated.

13. Electrical heating equipments are generally automatic, so it requires low attention and

supervision.

14. Protection against overheating can be provided by suitable switch gear.

b) 20 kW, 220 V resistance oven uses Nickel Chromium wire. If the temp. of charge is 727 °C and it is to be heated to 1127 °C, find the suitable length and diameter of wire. Assume : Emissivity = 0.9, Radiant efficiency = 0.6 & Sp. resistance = 1.03 x 10-6 SZ m.

Ans: (When Final answer of Numerical is correct Give Full Marks & if final answer is wrong give stepwise marks)

Given Data: T1 = 11270C = 1127+273 = 14000K ------------------------------------------- (1/2 Mark)

T2 = 7270C = 727 +273 = 10000K -----------------------------------------------(1/2 Mark)

Radiation efficiency = 0.6, specific resistance of Ni-Cr = 1.03x10-6 ohm m, emissivity = 0.9.

242414 m/w)

1000T()

1000T([e.k1072.5H

OR

24241 /])

100()

100([.72.5 mwTTek H

----------------------------- (1 Mark)

4 4 21400 10005.72 0.6 0.9 [ ( ) ( ) ] /

100 100w m H

287771.3408 /w mH

-------------------------------------------------------- (1 Mark)

PV

dl

4

2

2 ---------- Equation No.1----------------------------- (1 Mark)





2

2 6

(220)4 20 1000 1.03 10

ld

2 1845543.68l

d

21845543.68l d --- Equation No.2 ----------------------------- (1 Mark)

2 2

6

2 2

62

6 2 2

7 4

74

73

3 12

4

4 1.03 10 87771.34(220)

7.471

7.474 10 [1845543.678 ]2.5446 10

2.5446 10

1 2.5446 10

39296.5 10

d Hl Vdldld dd ddd

dd

Taking Cube root of both sides 33.399 10

3.399d md mm

--- ----------- (1Mark)

Substitute Value of ‘d’ in Equation No.2 to calculate ‘l’ :

21845543.68l d ----------------------------- (1 Mark)

3 21845543.68 [3.399 10 ]l

21.209172l m ------------------------------------------------- (1 Mark)

: 21.209172Answer Length l mtr 3.399Diameter d mm





c) What is electric welding ? Describe electric arc welding in brief. How arc is formed in electric arc welding ?

Ans: Meaning of electric welding : ( 2 Marks)

It is the process of joining two similar or dis-similar metals by application of heat with or

without application of pressure and addition of filler material.

Define electric arc welding:- ( 2 Mark )

The processes in which two metal parts to be welded are brought to a molten state and then

allowed to solidify is called as arc welding or stick welding.

How arc is formed :- for following method

a) By applying High Voltage

b) By separation of two current carrying electrodes suddenly

Explanation:- (Any one Method is expected 4 Marks) ( 4 Marks )

a) By applying High Voltage:- Figure:

or equivalent figure Operation:

When very high voltage is applied across any two electrodes separated by small air gap then air

between two electrodes gets ionized and ionized air is conducting, so current starts flowing from

one electrode to another electrode in the form of spark (arc).

This arc produces heat energy which is utilized for melting the charge.

High Voltage is required to produce arc and to maintain arc high voltage is not necessary.

Once arc is struck between two electrodes then low voltage is sufficient to maintain the arc.





b) By Separation of two current carrying electrodes suddenly:- Figure:

or equivalent figure

Operation:-

Another way to produce arc is to short circuit two current carrying electrodes as shown in fig (a)

and suddenly withdraw them, then there will be spark between two electrodes as shown in figure

(b)

This arc then produce heat energy which is utilized for melting the charge.

In this method high voltage is not necessary to produce the arc. Characteristics of Arc:

1. Arc is conducting. 2. Arc has negative temperature coefficient of resistance.

Q.4A) Attempt any THREE : 3 x 4 = 12 Marks

a) Compare DC and AC welding on any four points. Ans: ( Any Four point Expected: 1 Mark each, Total 4 Marks.)

S.No Points DC Welding AC Welding

1 Supply equipment used DC differential Compound Generator, or Rectifier

Welding Transformer

2 Heating Effect Uniform Not Uniform 3 Temperature Obtain More Less 4 Possibility of Arc Blow More Possibility No Possibility

5 Stability of Arc D.C Differential compound. Generator has dropping characteristics.

Use of series Reactor





6 Type of Electrode Non Coated Electrode is used Coated Electrode is compulsory

7 Voltage Required 50 to 60 volt 72 to 100 volt

8 Capital Cost High Low 9 Running cost High Low

10 Maintenance cost High Low 11 Stand by losses High by 25% Low 12 Efficiency Low, 65% High, 85% 13 Application Carbon Arc Welding Resistance Welding , Metal Arc

Welding

b) Describe with neat sketch operation of seam type resistance welding.

Ans:

1) Seam Welding its neat labelled sketch: ( 2 Mark)

or equivalent figure

Explanation: ( 2 Mark) Seam welding is nothing but series of continuous spot welding

Working:

Job is kept in between two electrodes under pressure. This pressure is kept constant throughout.

In this type intermittent current is used, it means current is ON for definite time and OFF for

another time interval with the help of timer.

If current is continuously passes then heat produced may cause burning of job.

Heat is produced due to I2R losses where ‘R’ is the contact resistance.




Subject Code: 17507 Model Answer of 42 This heat is utilized to obtain welding temperature (to become a plastic state)

When welding temperature is reached supply is cut down and external pressure is applied simultaneously across the job to complete weld.

c) Describe the construction of high pressure mercury vapour lamp with neat sketch, Ans: Figure mercury vapour discharge lamp :- (2 Mark)

Construction:- (2 Mark) It consists of an inner bulb generally of silicon, to withstand high temperatures.

The bulb contains a small quantity of mercury and argon.

It is protected by outer glass, this may be cylindrical or elliptical.

The space between the two bulbs is filled with nitrogen at a pressure of half atmosphere.

The discharge tube has three electrodes, namely two main electrodes A and B and one starting

electrode.

The starting electrodes are connected through a resistance of about 10-30 k ohm to the main

electrode, located at the far end.

The electrodes are of tungsten wire helices filled with electron emissive materials, usually barium

and strontium carbonates mixed with thorium.

OR Student may write

The construction & connection diagram is as shown in figure. As per this

construction there are following components.

Choke: The choke is acting as the ballast. At the time of supply voltage variation of current





flowing through the inner tube is maintained constant to keep uniform light intensity. Sometimes

choke can be designed for to get the higher voltages & to apply the inner tube of mercury vapour

lamp.

Starting resistance/limiting resistance: Whenever current flows through the starting resistance

there is a I2R loss which is converted into heat. If the temperature of this heat goes near about

6000C then there will be heating effect & inert gases ionization will be start.

Auxiliary electrode & Main electrode: It is made by high resistive element. The ionization is

taking place through the inert gases whenever current flows from auxiliary electrode to main

electrode.

Inner Tube: The various inert gases e.g. Argon, Nitrogen etc with mercury powder are filled in

the inner tube at low pressure or high pressure.

Outer Tube: The function of outer tube is to make the vacuum surrounding the inner tube to

avoid thermal dissipation or to maintain 6000C surrounding the inner tube.

Power factor improvement Capacitor: The function of power factor improvement capacitor is

to improve the power factor 0.5 to 0.95

d) Give the two laws of illumination.

Ans: (Inverse Square Law :- 2 Marks , Lamberts Cosine Law:- 2 Marks , Total 4 Marks.) 1) Inverse Square Law:-

Intensity of illumination produced by a point source varies inversely as square of the distance

from source.

2dIE

Where, I = intensity and d = Distance





2) Lamberts Cosine Law: According to this law, Illumination at any point on a surface is proportional to the

cosine of the angle between the normal at that point and the direction of luminous flux

Q. 4 B) Attempt any ONE of the following : 6 Marks

a) Describe through illustration the following types of lighting scheme : (i) Direct, (ii) Indirect, (iii) Semi-direct, (iv) Semi-indirect.

Ans: 1. Direct lighting: (1.5 Marks)

Application: The direct lighting scheme is widely used in drawing room, workshop and flood lighting etc. 2. Indirect lighting: (1.5 Marks)

Application:

Which is useful for drawing offices and composing rooms. It is also used for decoration

purposes in cinema halls, hotels etc.





3. Semi direct lighting: (1.5 Marks)

Application:

It is mainly used for interior decoration.

4. Semi indirect lighting: (1.5 Marks)

or equivalent figure.

Application:

It is mainly used for interior decoration.

b) Describe with schematic diagram steps involved in series — parallel • control of traction motor. Ans: (Series steps--- 3 Marks, Parallel steps----- 3 Marks, Total 6 Marks)

Series parallel control of DC series motor

1. For traction purpose, two motors are operated in following steps.

Series steps of traction motor:

Step 1 – Two traction motors M1 and M2 are connected in series and started with all starting resistances in

series.





The starting resistances are cut out one by one gradually from step 1 to step 7 and finally two motors are in series without any resistance.

In series connection the supply voltage V is divided in two motors. (Both motors get half or (V/2) volts). So speed is also half. (N/2)

Parallel steps of traction motor:

Step 1 –

After completion of series last step motors are now connected in parallel again with series resistance otherwise motor will draw very high current and may damage itself.

Step 2 to 7 –

Both motors are now connected in complete parallel and starting resistances are cut out one by one 2 To 7

In parallel connection, voltage across M1 and M2 will be full i.e. V (voltage is always same in parallel).

Voltage across each motor = V and speed of each motor = N

So, voltage is now increased from (V/2) to V. Hence, speed also increases from (N/2) to N and motor runs with full speed.





Advantages:

1. This method has highest starting efficiency then rheostat method.

Starting efficiency of plain rheostat method = 50 %. By this method for two motor it is

66.66% & for 4 motors it is 72.72% and for 6 motors it is 75%

2. Different economical speeds are obtained:

For 2 Motor = 1 :2

For 4 Motor = 1:2:4

For 6 Motor = 1:2:3

3) For same power input torque of different magnitude is obtained.

Disadvantages:

1. If proper transition method is not used then

There is loss of torque when motors are disconnected from supply

There will be jerk when motors are reconnected in parallel

Q.5 Attempt any FOUR : 4 x 4 = 16 Marks a) Write different systems of track electrification.

Ans: (Any Four Systems Of Track Electrification From The Following Are Expected 1 Mark

To Each Systems Of Track Electrification, Total 4 Marks) Following are the different track electrification system

D.C. Supply system:-

1. Direct current track electrification:

600V, 750V DC for tramways





1500V, 3000V DC for Train (Urban and sub-urban services)

A.C. Supply system:-

2. 1-Ph, 25KV,standard frequency AC supply system:

1-Ph, 25 KV, 50 Hz

3. 1-Phase, low frequency AC Supply system:

1-Ph, 15/16 KV, 16.2/3 Hz or 25 Hz

4. 3-Ph, Low frequency AC supply system;

3-Ph, 3.3/3.7 KV, 16 2/3 Hz or 25 Hz

Composite system:-

5. 1-Ph AC (1-ph, 25KV) – DC Supply System

6. Kando System (1-Ph AC – 3-Ph AC)

b) Write eight desirable characteristics of traction motor. Ans: (Any Eight Points From The Following Or Equivalent Points Expected 1/2 Mark Each,

Total 4 Marks) Desirable characteristics of ideal traction motors:-

1) It should be robust in construction to withstand against continuous vibrations.

2) Weight of motor per HP should be minimum in order to increase pay load capacity.

3) It must be small in overall dimensions, especially in overall diameter.

4) It must have totally enclosed type enclosure to provide protection against entry of dirt, dust, mud,

water etc. in drive.

5) When motors are running in parallel they should share almost equal load. (even when there is

unequal wear & tear of driving wheels)

6) It should have high starting torque.

7) It should possess high rate of acceleration & retardation.

8) It should be variable speed motor.

9) Its speed-torque characteristics should be such that it should produce high torque at low speed





and low toque at high speed.

10) Motor must be capable of taking excessive overload in case of emergency.

11) It should have simple speed control methods.

12) Electrical braking system should be reliable, easy to operate and control, especially regenerative

braking is possible.

13) Motor should draw low inrush current (Starting current,and if supply is interrupted and restore again.)

14) It should withstand for voltage fluctuation without affecting its performance. 15) It should have low initial cost. 16) It should have less maintenance cost. 17) It should have high efficiency.

18) It should have long life.

c) A train has schedule speed of 60 kmph between stops which are 6 km apart. Determine crest speed over the run assuming : (i) Duration of stops as 60 sec. (ii) Acceleration as 2 kmphps (iii) Retardation as 3 kmphps. The speed time curve is trapezoidal.

Ans: Given data: 60 / 6 50sec 2 / sec 3 / .secsch StopV Km hr D KM T km hr km hr

Solution;

3600( )sch

sch

DVSchedule Time T

------------------------------------------------------ (1 /2 Mark )

3600( )schsch

DSchedule Time TV

3600 6( )60schSchedule Time T

21600( )60schSchedule Time T

( ) 360 secschSchedule Time T

------------------------------------------------------- (1 /2 Mark )

( ) ( ) ( )sch stopSchedule Time T Actual Time of Run T Stop time T

( ) ( ) ( )sch stopActual Time of Run T Schedule Time T Stop time T

( ) 360 60Actual Time of Run T

( ) 300 secActual Time of Run T -------- ------------------------------------ ( 1/2 Mark)




Subject Code: 17507 Model Answer of 42 Maximum Speed

KDKTTV

2360042

max

But,

2

K- ------------- ------------------------------------------------ (1/2 Mark )

2 3

2 2 3K

K = 0.4167 ------------ ----------------------------------------------- (1/2 Mark )

Now,

KDKTTV

2360042

max

- ----------------------------------------- (1/2 Mark )

2

max300 (300) 4 0.4167 3600 6

2 0.4167V

max

300 232.3722 0.4167

V

max 81.1471 /V Km hr

-- ------------------------------------------------------------( 1 Mark)

d) Draw a neat labelled block diagram of AC electric locomotive. State the function of each part. Ans: labelled diagram of AC electric locomotive: ( Diagram: 2 Marks & Function: 2 Marks)





Function of AC electric locomotive Parts:

1) Overhead contact wire:

Supply of 1-ph, 25KV, 50Hz, AC is given to overhead conductor.

2) Current collecting device:

It collects current from overhead contact wire and passes it to tap changing transformer through

circuit breaker.

3) Circuit breaker (C.B):

It is connected in between current collecting devices and tap changing transformer.

SF6 circuit breaker is used.

To disconnect locomotive equipments whenever there is fault.

It opens automatically when train passes neutral zone (from zone No.1 to Zone No.2)

4) On load tap changing transformer:

It changes the tap without disconnecting the load on transformer. Its purpose is to vary the

voltage for speed control of traction motor.

5) Traction Transformer:

It step down input voltage 25 KV to working voltage of traction motor (1500V/3000V).

6) Rectifier:

It converts secondary voltage of transformer into DC supply.

7) Filter circuit (smoothing reactor):

It is used to obtain pure DC supply.

8) Motor control unit: It controls operation of traction motor.

9) Traction Motor:

It gives mechanical power to run the train i.e. DC series motor which is used as traction

motor.





e) "DC series motor is used for traction purpose." Justify your answer with any six characteristics. Ans: ( 4 Marks)

Due to following characteristics and advantages, DC series motor is suitable for traction purpose: 1) Characteristics: We know that,

Due To Following Reasons DC Series Motor Is Used For Traction Purpose:-

1. DC Series motor robust in construction and capable to withstand against continuous vibration.

2. DC series motor weight is 1.5 times less than 1-Ph AC series motor for same H.P.

3. DC Series motor has high starting torque.

4. DC Series motor has high rate of acceleration and retardation.

5. DC Series motor is variable speed motor. Due to these characteristics motor is protected against

overload.

6. DC Series motor speed-torque characteristics are such that as torque increases speed decreases.

7. DC series motor has develops high torque at low speeds, low torque at high speed, this is the basic requirement of traction unit.

8. Commutating property of series motor is good so we get sparkles commutation.

9. Torque is unaffected by variation in supply voltage.

10. DC Series motor maintenance cost is less.

11. When DC series motor are running in parallel the all motors share almost equal load.

12. Torque obtained by DC series motor is smooth and uniform, so it improves riding quality.





e) Draw speed time curve. Show and list various time periods associated with it. Ans: Typical speed time curve for main traction line :

( Total, 4 Mark)

OR

Speed Time Curve list various time periods:-

There are five periods in the run of train as shown in speed time curve.

i) Constant acceleration period (o to A)

ii) Acceleration on speed –Time curve (A to B) For T2 sec.

iii) Free Running or constant period (B to C) For T3 sec.

iv) Coasting period (C to D) For T4 sec.

v) Braking period (D to E) For T5 sec.





Q.6 Attempt any TWO : 8 x 2 = 16 Marks

a) i) A 400 V, 50 Hz, 3 Phase line delivers 200 kW at 0.7 p.f. lagging. It is desirable to improve the line power factor to unity by using shunt capacitors. Calculate value of capacitance of each unit if they are connected in delta.

Ans: Volt : line volts V = 400V, f= 50 Hz P= 200kW cos 1 =0.7 cos 2 =1

011 5729.457.0Cos

01 5729.42tantan

tan 1 = 1.020 ----------------------------------------------------------- (1/2 Mark) tan 2 = 0 ------------------------------------------------------------- (1/2 Mark) Q1 = P tan 1 = 200 x 1.020 = 204 KVAR ----------------------------------------------------------- (1/2 Mark)

Q2= P tan 2 = 200 x 0 = 0 KVAR ----------------------------------------------------------- (1/2 Mark) QC = Q1- Q2

= P tan 1 - P tan 2 --------------------(1/2 Mark)

= 204-0

= 204 KVAR ---------------------------------------------------- (1/2 Mark)

Capacitor when connected in Delta:-

2C

V3QphaseperC

------------------------------------------------------- (1/2 Mark)

2

3

400502310204phaseperC

6

3

10265.50310204phaseperC

3103528.1phaseperC F ------------------------------------------------ (1/2 Mark)





a) ii) State four requirements of tariff.

Ans:

(Any Four Requirements From The Following Or Equivalent Points Are Expected , Total 4 Marks) Following are the requirements of Tariff :-

1. It should be easy to understand to consumer.

2. Easy to calculate.

3. Tariff should be attractive i.e. It should not be too high or too low. It should be reasonable.

4. Tariff should be economical as compare to other types of energy sources.

5. Tariff should be different for different types of consumers.

6. Tariff must be fair, so that different types of consumers are satisfied with rate of electrical energy charges.

7. Tariff should be framed into two parts i.e. fixed charges + running charges.

8. Tariff should be high during peak load period.

9. Tariff should be low during off load period.

10. For industrial consumer, in addition to basic tariff incentives and penalty related to

PF and LF should be considered.

b) (i) What are different tariffs used by electricity supply authority ? Describe any two in brief. Ans: Types of Tariff:- ( Any Four Types expected: 1/2 each, Total 2 Marks)

1) Flat-demand Tariff

2) Simple-demand Tariff or Uniform Tariff

3) Flat-rate Tariff

4) Step-rate Tariff

5) Block-rate Tariff

6) Two-part Tariff

7) Maximum demand Tariff

8) Three-part Tariff





9) Power factor Tariff :- a) KVA maximum demand Tariff

b) Sliding Scale Tariff or Average P.F. Tariff

c) KW and KVAR Tariff

10) TOD (Time of Day) Tariff

11) ABT:-This tariff system is called availability based tariff. As its name suggest it is a tariff

system which depends on the availability of power.

Explanation of Types of Tariff ( Any TWO Types explanation Expected: 1 Mark each,

Total 2 Marks)

1) Block Rate Tariff:-

In case of block rate tariff there are blocks of units consumed and each block tariff rate/unit

(KWH) is different plus consumer has to pay fix charges e.g.

If generation is less than utilization than tariff rate/unit in each block goes on increasing and

vice versa. e.g.

2) Two Part Tariff:-

In this type of tariff energy bill is split into two parts.

ENERGY BILL= FIXED CHARGE which depends on load (KW)

+RUNNING CHARGE which depends on actual energy consume (KWH)

Fixed charge which depends on load (KW) which is declared by consumer on test report.

There is no separate meter is installed to measure load.

Only one energy meter is used to measure number of units consumed.

This type of tariff system is used for residential and commercial consumers.(up to 20 KW)

This type of tariff is not used for industrial consumers.

Advantages:

1. It recovers fixed charges which depends on load (KW), so it automatically recovers capital investment of Supply Company





Disadvantages:

1. The consumer has to payfix charges per month whether he has to consume or not consume the electrical energy.

Application:

1. This type of tariff system is used for residential and commercial consumers. (Up to 20 KW)

2. This type of tariff is not used for industrial consumers.

3) Maximum Demand Tariff/KVA Maximum Demand Tariff / Load factor tariff:-

This is basic tariff for all industrial / commercial consumers with contract demand above 80

KW/ 100KVA/107 HP

It is similar to two part tariff except that maximum demand (KVA) is actually measured by

installing maximum demand meter(in KVA)

M.D. Meter (it is an electromagnetic or electronic trivector meter) is installed in the premises

of consumer, in addition to energy meter.

Maximum Demand Tariff / Load factor Tariff =

"'})({'')(.. YRsconsumerActualKWHunitsofNumberpermonthXRsKVADM

Application: -This type of tariff is applicable to industrial consumer/H.T/ commercial consumers with

contract demand above 80 kw/ 100Kva/107 hp consumer.

Measurement of KVA M.D.:-

Actual Maximum Demand recorded in the month during 06am.To10pm. Is considered for billing.

Incentives and Penalties to M.D. tarrif :- Incentives :-

1) If consumer is used M.D. above 75 % to 85 % of saction contract demand than , consumer will gate 0.75 % rebeat on the energy bill.

2) If consumer is used M.D. above 85 % to 100 % of saction contract demand than , consumer will gate 1 % rebeat on the energy bill.

Penalties :- 1) If consumer is used M.D. above 100 % of saction contract demand than , consumer has to pay





more demand charges 150 % for use of extra M.D.

2) If consumer is used M.D. below 50 % of saction contract demand than , consumer has to pay

minimum demand charges 50 % of saction contract demand.

Advantages:-

1) Each industrial consumer is trying to use M.D. above 75 % to 100 % of sanction contract

demand to get discount in energy bill. So it will improve load factor of industry.

2) Industrial consumers were not utilizing their load simultaneously to avoid Penalties on exceed

of M.D. than saction contract demand. So it will improve diversity factor.

3) Industrial consumer is trying to improve power factor to reduce maximum demand charges.

Since pfIKVA 1

4) As each industry run at high load factor, diversity factor and power factor then overall load

factor, diversity factor and power factor of power system increases.

5) Which will automatically beneficial from the economics of power system and energy

conservation point of view.

4) Power Factor Tariff:-

In addition to basic tariff (Maximum Demand Tariff/KVA Maximum Demand Tariff / Load factor tariff) the tariff in which P.F. of industrial consumer is taken into consideration.Is known as Power Factor Tariff.

If the P.F. of consumer is less than P.F. declare by Supply Company (say below 0.9 Lag.) than

penalty will be charged in energy bill.

If The P.F. of consumer is more than P.F. declare by Supply Company (say above 0.95lag.)

than discount will be given in energy bill.

As usual consumer has to pay actual energy consumption charges

Application :-

This type of tariff is applicable to industrial consumer/H.T/ commercial consumers with






Incentives and Penalties to Power factor tarrif :- Power factor incentive:- e.g.

Power Factor Percentage of incentive

0.95 0% of energy bill

Above 0.96 1% of energy bill




At unity P.F. 5% of energy bill

Power factor penalty:- e.g.

Power factor lagging Percentage of penalty

For0.90Power factor lagging 0% of energy bill

For 0.89 Power factor lagging 2% of energy bill













There are three types of P.F. tariff ;-

a) KVA maximum demand Tariff: (All ready explain above)

b) Sliding Scale Tariff or Average P.F. Tariff:

If the P.F. of consumer is less than P.F. declare by Supply Company (say below

0.9 Lag.) than penalty will be charged in energy bill.

If The P.F. of consumer is more than P.F. declare by Supply Company (say above

0.95lag.) than discount will be given in energy bill.

As usual consumer has to pay actual energy consumption charges

c) KW and KVAR Tariff:

In this type both active (KW) & reactive power (KVAr) supplied are charged

separately and actual energy consumption charges

A consumer having low power factor draw more reactive power and shall have to

pay more charges and vice-versa.

So consumer is trying to improve power factor to reduce KVAr charges in energy

bill, so power factor of power system increases.

}arg)(''{}arg)(''{}arg)(''{ esChKWHCRsesChKVARBRsesChKWARsBillEnergy

5) Time of Day (TOD) Tariff or OFF-load Tariff:-

In addition to basic tariff (Maximum Demand Tariff / KVA Maximum Demand Tariff /

Load factor tariff also the tariff in which P.F. of industrial consumer is taken into consideration.)

Consumer has to pay energy consumption charges according to time for which energy is

consumed.

TOD energy meter is installed in the consumer premises.

This meter is specially designed to measure energy consumption w.r.t. time.

This type of tariff is such that energy consumption charges/unit are less at during OFF-load

period





Energy consumption charges/unit are more during PEAK -load period

This type of tariff is introduced to encourage industrial consumers to run their maximum load

during OFF-load period.

e.g.

Sr.No Block Rate / KWH Rs Remark

1 8.00 am to 12.00 noon Rs. 6.00 per unit+0.80 Rs. Per unit Peak load period

2 12.00 noon to 6.00 pm Rs. 5.00 per unit+ 0 Rs. Per unit Base load

3 6.00 pm to 10.00 pm Rs. 6.00 per unit+ 1.10 Rs. Per unit Peak load period

4 10.00 pm to 8.00 am Rs. 5.00 per unit – 1.50 Rs. Per unit OFF load period

Application :-

This type of tariff is applicable to industrial consumer/H.T/ commercial consumers with


6) Three part Tariff:-

Fixed charges per month depend on connected load.

Semi-fixed charges depend on KVA maximum demand.

Running charges depend on actual energy consume.

b) (ii) State any four advantages of good power factor for electric supply. Ans: Following Advantages of good power factor for electric supply:

( Any Four Advantages are expected: 1 Mark each, Total 4 Marks) 1. Cross section of conductor reduces:

Cross section of conductor f.P

1I

As P.F. increases current reduce so; cross section of conductor and its weight reduces hence its cost reduces

2. Design of supporting Structure:

As weight of conductor reduces design of supporting structure (tower) becomes lighter, so its cost reduces.





3. Cross section of terminal (contacts) reduces:

As power factor increases, current reduces. hence cross section of switchgear bus bar and contacts etc decreases.

4. Copper losses reduces:

As power factor increases current reduces. So copper losses reduces. As a effect efficiency increase.

5. Voltage drop reduces:

As P.F. increases, current decreases. So voltage drop decreases, So regulation gets improved (better)

6. Handling capacity (KW) of equipment increases:

As power factor increases, handling capacity of each equipment such as Alternator,

transformer increases

7. KVA rating of equipments reduces:

As P.F. increases, current decreases. So KVA rating of all equipments for eg- alternator, transformer etc decreases, so its capital cost reduces.

8. Cost per unit (KWH) reduces:

From all above advantages, it is seen that cost of generation, transmission & distribution decreases, so cost/unit reduces.

Also performance i.e. efficiency & regulation gets improved at high power factor

c) (i) A Factory takes 300 kW at 110 V from a 3 phase supply and power factor of 0.7 lagging. A

synchronous motor is installed which takes an additional 150 kW. What must be the kVA rating of this motor to raise the power factor of the system to 0.85 lagging ?

Ans:

or equivalent diagram Given Data: PL = 300 KW 0.7 0.7 1Cos lag Sin tan





Power factor improved to 0.85 lag tan 0.6197 150new m KWP

Reactive Power taken by load (QL) = PL tan ------------------------------------ (1/2 Mark)

= 300 x 1

= 300 KVAR (lag) -------------------------- (1/2 Mark) Reactive Power taken after synchronous motor is connected (Qnew) =

=(PL + Pm) tan new -------------------------- (1/2 Mark)

= (300 +150) x 06197 = 450 x 06197

= 278.8849 KVAR (lag) ------------------ (1/2 Mark)

Reactive Power taken by synchronous motor to improve P.f = = ( ( )L new

= 300 - 278.8849

= 21.115 KVAR ( leading) --------------- ( 1/2 Mark)

2( )m m mKVA Rating of Synchronous Motor S P - ---------------------- ( 1/2 Mark)

2 2(150) (21.115)mS

22945.845mS

151.4788mS KVA ----------------------- ( 1/2 Mark)

150151.4788

mm

m

PPower Factor of Synchronous Motor CosS

0.9902mPower Factor of Synchronous Motor Cos leading --------------------- ( 1/2 Mark)





c) (ii) Derive the equation of most economical power factor.

Ans: Derivation: ( 4 Mark)

Let,

P = Active power KW

S1, S2 = KVA Maximum demand before and after improving power factor

Q1, Q2 = Lagging reactive power before & after improving power factor

QC = Leading Reactive power drawn by Capacitor

1Cos = Initial Power factor

2Cos = Improved Power factor

Rs X = Tariff charges towards M.D. (KVA) /year

Rs Y = Expenditure towards KVAr to be neutralized per year (Expenditure towards P.F. improving apparatus)

1) Before improving Power factor:

11 tanPQ

11 S

PCos

11 Cos

PS





111 secP)S(KVA

2) After improving Power factor:

22 tanPQ

22 S

PCos

22 Cos

PS

222 secP)S(KVA

3) Saving in KVA charges:

= Rs X (S1 –S2)

= Rs X ( 21 secPsecP )

= Rs X .P ( 21 secsec )

4) Expenditure towards KVAr to be neutralized:

= Rs Y (Q1 –Q2)

= Rs Y ( 21 tanPtanP )

= Rs YxP ( 21 tantan )

5) Net Saving:

= Saving in KVA charges - Expenditure towards KVAr to be neutralized.

= [Rs X .P ( 21 secsec )] - [ Rs Y ( 21 tanPtanP )]

Saving will be maximum when differentiate above equation with respect to 2 and equate to zero.





)tan(tanPYRs)sec(secPXRsdd

dds

212122

22

22 secPY0tansecPX0

22

22 secPYRs0tan.secPXRs0

22

22 secPYRstan.secPXRs

22 secYRstanXRs

22

2

Cos1YRs

CossinXRs

YRssinXRs 2

XYRssin 2

6) 1sin 22

22 Cos

122 Cos 2

2sin

Most economical power factor = 22 )/(1 xYCos

Most economical power factor at which maximum saving will occurs

-------------------------------------------------- END---------------------------------------------------------------

any four point expected: 1/2 each point: 2 marks)

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